如何检查熊猫数据框中的卡住数据

Question

为了清楚起见，已根据 cmets 对这个问题进行了编辑

目标：基于组标记数据框列中可变数量的卡住/重复值。

所需的解决方案：所需的解决方案应满足以下条件：

1. 速度和简洁很重要
2. 解决方案应接受可变数量的阈值作为输入，用于标记每个区域的卡住/重复值。 例如: {"Zone1": 4 , "Zone2":2} ->表示在Zone1中，需要有4个连续的重复值才能被逻辑标记，而在Zone2中，至少有2个或更多的重复值应该触发标记
3. 请尽可能添加 cmets 以便于理解
4. 最好使用 Pandas 和/或 Numpy
5. 在一个包含 87600 个值的测试数据帧上为您的解决方案计时（生成一个包含 87600 个值的数据帧）

输入数据/问题表述：

import pandas as pd
import numpy as np
from random import randint

# Generate some random data
ts_index = pd.date_range("1/1/2019", periods=24, freq="1H")
v1 = [randint(1, 100) for i in range(24)]
v2 = [2] * 24
v3 = [2, 2, 2, 2, 4, 4, 0, 2, 2, 1, 9, 2, 4, 1, 2, 2, 0, 2, 1, 8, 1, 7, 3, 5]
test_df = pd.DataFrame({"v1": v1, "v2": v2, "v3": v3}, index=ts_index)
print(test_df) 

现在，示例输出应如下所示。

Answer 1

首先我们得到列的绝对差，然后是cumsum()。

假设如果一个值没有变化，则差异的累积和将保持不变，我们可以使用 pandas .shift()。

我们必须将 zone1 中的每个值与上面的 4 行进行比较。 zone2 类似。

import pandas as pd
import numpy as np
import datetime


# Generate some random data
ts_index = pd.date_range("1/1/2019", periods=24, freq="1H")
v1 = [random.randint(1, 100) for i in range(24)]
v2 = [2] * 24
v3 = [2, 2, 2, 2, 4, 4, 0, 2, 2, 1, 9, 2, 4, 1, 2, 2, 0, 2, 1, 8, 1, 7, 3, 5]
df = pd.DataFrame({"v1": v1, "v2": v2, "v3": v3}, index=ts_index)
df1 = df.diff().fillna(0)

# Add hour and time of day to create flag
df['Hour'] = df.index.hour
df['Flag'] = np.where((df['Hour'] <= 8) | (df['Hour'] >= 18), 'Zone1', 'Zone2')
df1['Flag'] = np.where((df['Hour'] <= 8) | (df['Hour'] >= 18), 'Zone1', 'Zone2')

df1[["v1", "v2","v3"]] = np.abs(df1[["v1", "v2","v3"]])
df1[["v1", "v2","v3"]] = df1.groupby("Flag")["v1", "v2","v3"].cumsum()
columns = ["v1","v2","v3"]

z1_p = 3
z2_p = 1

for col in columns:
    df["Flag_"+col] = (np.array(df1[col].shift(z1_p) == df1[col]) & np.array(df1["Flag"] == "Zone1")) | \
(np.array(df1[col].shift(z2_p) == df1[col]) & np.array(df1["Flag"] == "Zone2"))
    for element in df[df["Flag_"+col] == True].index:
        if df.loc[element]["Flag"] == "Zone1":
            for i in range(1,4):
                a = a.append(pd.Index([element - datetime.timedelta(hours=i)]))
        else:
            a = a.append(pd.Index([element - datetime.timedelta(hours=1)]))
    df.at[a,"Flag_"+col] = True

df

输出：

                    v1  v2  v3  Hour    Flag    Flag_v1 Flag_v2 Flag_v3
2019-01-01 00:00:00 31  2   2   0       Zone1   False   True    True
2019-01-01 01:00:00 93  2   2   1       Zone1   False   True    True
2019-01-01 02:00:00 48  2   2   2       Zone1   False   True    True
2019-01-01 03:00:00 56  2   2   3       Zone1   False   True    True
2019-01-01 04:00:00 9   2   4   4       Zone1   False   True    False
2019-01-01 05:00:00 75  2   4   5       Zone1   False   True    False
2019-01-01 06:00:00 29  2   0   6       Zone1   False   True    False
2019-01-01 07:00:00 61  2   2   7       Zone1   False   True    False
2019-01-01 08:00:00 64  2   2   8       Zone1   False   True    False
2019-01-01 09:00:00 82  2   1   9       Zone2   False   True    False
2019-01-01 10:00:00 13  2   9   10      Zone2   False   True    False
2019-01-01 11:00:00 97  2   2   11      Zone2   False   True    False
2019-01-01 12:00:00 74  2   4   12      Zone2   False   True    False
2019-01-01 13:00:00 26  2   1   13      Zone2   False   True    False
2019-01-01 14:00:00 77  2   2   14      Zone2   False   True    True
2019-01-01 15:00:00 39  2   2   15      Zone2   False   True    True
2019-01-01 16:00:00 79  2   0   16      Zone2   False   True    False
2019-01-01 17:00:00 35  2   2   17      Zone2   False   True    False
2019-01-01 18:00:00 65  2   1   18      Zone1   False   True    False
2019-01-01 19:00:00 74  2   8   19      Zone1   False   True    False
2019-01-01 20:00:00 72  2   1   20      Zone1   False   True    False
2019-01-01 21:00:00 23  2   7   21      Zone1   False   True    False
2019-01-01 22:00:00 28  2   3   22      Zone1   False   True    False
2019-01-01 23:00:00 59  2   5   23      Zone1   False   True    False

Answer 2

这行得通吗

import pandas as pd, numpy as np
from random import randint


def processGroup(zone, ZoneGroup):
    ZoneGroup.sort_index(inplace=True)
    repeatation = repeatations[zone]
    
    repeat_counts = [0]*len(cols)
    previous_value = [None]*len(cols)
    
    for i, record in ZoneGroup.iterrows():
        for j, c in enumerate(cols):
            if previous_value[j] == record[c]:
                repeat_counts[j] += 1
            else:
                repeat_counts[j] = 0
                
            if repeat_counts[j] >= repeatation:
                test_df.at[i, c+'RepeatFlag']=True
                
            previous_value[j] = record[c]

TotalRecords = 87600 #24*10
ts_index = pd.date_range("1/1/2019", periods=TotalRecords, freq="1H")
v1 = [randint(1, 2) for i in range(TotalRecords)]
v2 = [randint(1, 3) for i in range(TotalRecords)]
v3 = [randint(1, 5) for i in range(TotalRecords)]

test_df = pd.DataFrame({"v1": v1, "v2": v2, "v3": v3}, index=ts_index)
test_df['Zone'] = pd.Series(test_df.index).apply(lambda t: 'Zone'+ str(int((t.hour+t.minute/60+t.second/3600)// 8 + 1))).to_list()

# You can tweak repaet values here
repeatations = {'Zone1':4, 'Zone2':3, 'Zone3':2}
cols = test_df.columns[:-1]

#test_df[cols+'RepeatFlag']=False
for c in cols+'RepeatFlag':
    test_df[c]=False

for i, g in test_df.groupby('Zone'):
    processGroup(i, g)
    
print(test_df)
repeat_df = test_df[test_df.v1RepeatFlag | test_df.v2RepeatFlag | test_df.v3RepeatFlag]
print(repeat_df)

Answer 3

我已经测试了代码。查看笔记本的屏幕截图。尝试一次编辑的代码。